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Chapter 1: Problem 59

In a reaction, \(34.0 \mathrm{g}\) of chromium(III) oxide reacts with \(12.1 \mathrm{g}\) of aluminum to produce chromium and aluminum oxide. If \(23.3 \mathrm{g}\) of chromium is produced, what mass of aluminum oxide is produced? )

Short Answer

Expert verified
In the given reaction, \(34.0 \mathrm{g}\) of chromium(III) oxide reacts with \(12.1 \mathrm{g}\) of aluminum and produces \(23.3 \mathrm{g}\) of chromium. By calculating the number of moles of each substance and using stoichiometry, we can determine that \(34.3 \mathrm{g}\) of aluminum oxide is produced.

Step by step solution

01

Write the balanced chemical equation for the reaction

First, we need to write a balanced chemical equation for the reaction between chromium(III) oxide and aluminum to produce chromium and aluminum oxide: \[2Cr_2O_3 + 3Al \rightarrow 4Cr + 3Al_2O_3\]
02

Calculate the number of moles of reactants and chromium produced

Now, we need to convert the mass of each reactant and the mass of chromium produced to moles. We will use the molar mass of each compound for this conversion: Molar Masses: - Chromium(III) oxide: \(Cr_2O_3 = 2 * 51.9961 \mathrm{g/mol} (Cr) + 3 * 15.9994 \mathrm{g/mol} (O) = 151.9904 \mathrm{g/mol}\) - Aluminum: \(Al = 26.9815 \mathrm{g/mol}\) - Chromium: \(Cr = 51.9961 \mathrm{g/mol}\) Moles of each substance: - Chromium(III) oxide: \(\frac{34.0 \mathrm{g}}{151.9904 \mathrm{g/mol}} = 0.2236 \mathrm{mol}\) - Aluminum: \(\frac{12.1 \mathrm{g}}{26.9815 \mathrm{g/mol}} = 0.4486 \mathrm{mol}\) - Chromium: \(\frac{23.3 \mathrm{g}}{51.9961 \mathrm{g/mol}} = 0.4481 \mathrm{mol}\)
03

Use stoichiometry to determine the number of moles of aluminum oxide produced

From the balanced chemical equation (Step 1), we can determine the stoichiometric relationship between the reactants and product: \[2Cr_2O_3 + 3Al \rightarrow 4Cr + 3Al_2O_3\] Using the moles of chromium produced (0.4481 mol), we can determine the number of moles of aluminum oxide produced: \[\frac{3 \mathrm{mol \, Al_2O_3}}{4 \mathrm{mol \, Cr}} * 0.4481 \mathrm{mol \, Cr} = 0.3361 \mathrm{mol \, Al_2O_3}\]
04

Calculate the mass of aluminum oxide produced

Finally, we can convert the number of moles of aluminum oxide produced to mass using its molar mass: - Aluminum oxide: \(Al_2O_3 = 2 * 26.9815 \mathrm{g/mol} (Al) + 3 * 15.9994 \mathrm{g/mol} (O) = 101.9612 \mathrm{g/mol}\) Mass of aluminum oxide produced: \(0.3361 \mathrm{mol \,Al_2O_3} * 101.9612 \mathrm{g/mol} = 34.3 \mathrm{g}\) So, 34.3 g of aluminum oxide is produced.

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